A sample of 4.5 mg of an unknown monohydric alcohol, R-OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is __________ g/mol. [Nearest integer]
Answer (integer)
33
Solution
$$\mathrm{R}-\mathrm{OH}+\mathrm{CH}_{3} \mathrm{Mgl} \Rightarrow \mathrm{R}-\mathrm{OMgl}+\mathrm{CH}_{4}$$<br/><br/>
moles of alcohol $(\mathrm{ROH}) \equiv$ moles of $\mathrm{CH}_{4}$<br/><br/> At STP,
[Assuming STP]
<br/><br/>
1 mole corresponds to $22.7 \mathrm{~L}$
<br/><br/>
Hence, $3.1 \mathrm{~mL} \equiv \frac{3.1}{22700} \mathrm{~mol}$
<br/><br/>
So, moles of alcohol $=\frac{3.1}{22700}$
<br/><br/>
$\Rightarrow \frac{3.1}{22700}=\frac{4.5 \times 10^{-3}}{\mathrm{M}}$
<br/><br/>
$M \simeq 33 \mathrm{~g} / \mathrm{mol}$
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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